Problem: Simplify and expand the following expression: $ \dfrac{3}{4p + 32}+ \dfrac{3}{3p - 24}+ \dfrac{p}{p^2 - 64} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{3}{4p + 32} = \dfrac{3}{4(p + 8)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{3}{3p - 24} = \dfrac{3}{3(p - 8)}$ We can factor the quadratic in the third term: $ \dfrac{p}{p^2 - 64} = \dfrac{p}{(p + 8)(p - 8)}$ Now we have: $ \dfrac{3}{4(p + 8)}+ \dfrac{3}{3(p - 8)}+ \dfrac{p}{(p + 8)(p - 8)} $ The least common multiple of the denominators is: $ 12(p + 8)(p - 8)$ In order to get the first term over $12(p + 8)(p - 8)$ , multiply by $\dfrac{3(p - 8)}{3(p - 8)}$ $ \dfrac{3}{4(p + 8)} \times \dfrac{3(p - 8)}{3(p - 8)} = \dfrac{9(p - 8)}{12(p + 8)(p - 8)} $ In order to get the second term over $12(p + 8)(p - 8)$ , multiply by $\dfrac{4(p + 8)}{4(p + 8)}$ $ \dfrac{3}{3(p - 8)} \times \dfrac{4(p + 8)}{4(p + 8)} = \dfrac{12(p + 8)}{12(p + 8)(p - 8)} $ In order to get the third term over $12(p + 8)(p - 8)$ , multiply by $\dfrac{12}{12}$ $ \dfrac{p}{(p + 8)(p - 8)} \times \dfrac{12}{12} = \dfrac{12p}{12(p + 8)(p - 8)} $ Now we have: $ \dfrac{9(p - 8)}{12(p + 8)(p - 8)} + \dfrac{12(p + 8)}{12(p + 8)(p - 8)} + \dfrac{12p}{12(p + 8)(p - 8)} $ $ = \dfrac{ 9(p - 8) + 12(p + 8) + 12p} {12(p + 8)(p - 8)} $ Expand: $ = \dfrac{9p - 72 + 12p + 96 + 12p}{12p^2 - 768} $ $ = \dfrac{33p + 24}{12p^2 - 768}$ Simplify: $ = \dfrac{11p + 8}{4p^2 - 256}$